June 2023 Paper 2 Q15
15 In this question you must show detailed reasoning.
The equation of a curve is
\(\ln y + x^3y = 8\).
Find the equation of the normal to the curve at the point where \(y = 1\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are constants to be found. [7]
| Scheme | Marks | AO |
|---|---|---|
| \(y = 1\) then \(x = 2\) only | B1 | 3.1a |
| \(\dfrac{1}{y} \times \dfrac{\mathrm{d}y}{\mathrm{d}x}\) | B1 | 2.1 |
| \(x^3 \times \dfrac{\mathrm{d}y}{\mathrm{d}x} + 3x^2y\) | M1 | 1.1 |
| \(\dfrac{1}{y} \times \dfrac{\mathrm{d}y}{\mathrm{d}x} + x^3 \times \dfrac{\mathrm{d}y}{\mathrm{d}x} + 3x^2y\ [= 0]\) | A1 | 1.1 |
| substitution of their \(x = 2\) and \(y = 1\) to obtain numerical value for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1* | 1.1 |
| \(y - 1 = \left(\textit{their } \frac{3}{4}\right)(x - \textit{their } 2)\) oe | M1dep* | 3.1a |
| \(3x - 4y - 2 = 0\) or \(-3x + 4y + 2 = 0\) oe | A1 | 3.2a |
| [7] |
Notes
B1: first term correct; allow \(y^{\prime}\) for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: Product Rule; allow one coefficient error or one index error
M1*: NB \(-\frac{4}{3}\)
dependent on at least two of 3 terms correct on LHS following differentiation;
if expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) or evaluation of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) is incorrect, need to see substitution for award of M1
M1dep*: FT negative reciprocal of their \(-\frac{4}{3}\) and their 2
may see eg \(1 = \frac{3}{4} \times 2 + c\)
A1: must be in required form, but coefficients may be fractions